Some Basic Concepts Of Chemistry Question 87
Question: If potassium chlorate is 80% pure, then 48 g of oxygen would be produced from (atomic mass of K=39)
Options:
A) 153.12 g of $ KClO _3 $
B) 122.5 g of $ KClO _3 $
C) 245 g of $ KClO _3 $
D) 98 g of $ KClO _3 $
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Answer:
Correct Answer: A
Solution:
[a] $ \underset{\begin{smallmatrix} 2\times 122.5g \\ ,245 \end{smallmatrix}}{\mathop{2KClO _3}},\xrightarrow{^{heat}}2KCl+\underset{\begin{smallmatrix} 3\times 32g \\ ,96 \end{smallmatrix}}{\mathop{3O _2}}, $ 48 g of oxygen will be produced from 122.5 g of $ KClO _3 $
$ \therefore $ Amount of 80% $ KClO _3 $ needed $ =\frac{100}{80}\times 122.5=153.12g $
